Question

A student is standing with a banner at the top of a 100 m high college building. From a point on the ground, the angle of elevation of the top of the student is $60^{\circ$ and from the same point, the angle of elevation of the top of the tower is $45^{\circ}$. Find the height of the student ?}

• A. 35.7 m
• B. 75.2 m
• C. 50 m
• D. 73.2 m
• E. 75 m

Hint:

In height and distance problems involving two objects stacked vertically, calculate the base distance $d$ first using the known height, then substitute it into the tangent equation for the combined height.

Answer 1

The Correct Option is D

Step 1: Setup the geometric relationships.
Let the distance from the observation point to the base of the building be $d$.
Height of the building (tower) = 100 m.
Let the height of the student be $h$. Total height to the top of the student = $100 + h$.

Step 2: Use trigonometric ratios.
From the angle of elevation to the top of the building:
$\tan(45^{\circ}) = \frac{100}{d} \Rightarrow 1 = \frac{100}{d} \Rightarrow d = 100$ m.

Step 3: Solve for the student's height.
From the angle of elevation to the top of the student:
$\tan(60^{\circ}) = \frac{100 + h}{d} \Rightarrow \sqrt{3} = \frac{100 + h}{100}$
$100\sqrt{3} = 100 + h \Rightarrow h = 100(\sqrt{3} - 1)$
Taking $\sqrt{3} \approx 1.732$, $h = 100(1.732 - 1) = 100(0.732) = 73.2$ m.